Chapter 28 Problem 61

Problem

Helmholtz coils are two identical circular coils having the same radius $R$ and the same number of turns $N$, separated by a distance equal to the radius $R$ and carrying the same current $I$ in the same direction. They are used in scientific instruments to generate nearly uniform magnetic fields.

(a) Determine the magnetic field $B$ at points $x$ along the line joining their centers. Let $x=0$ at the center of one coil, and $x=R$ at the center of the other.

(b) Show that the field midway between the coils is particularly uniform by showing ${\frac {dB}{dx}}=0$ and ${\frac {d^{2}B}{dx^{2}}}=0$ at the midpoint between the coils.

(c) If $R=10.0$ cm, $N=250$ turns and $I=2.0$ A, what is the field at the midpoint between the coils, $x=R/2$?

Solution

(a)

Choose $x=0$ at the center of one coil. The center of the other coil will then be at $x=R$.

$B(x)={\frac {\mu _{0}NIR^{2}}{2\left[R^{2}+x^{2}\right]^{3/2}}}+{\frac {\mu _{0}NIR^{2}}{2\left[R^{2}+(x-R)^{2}\right]^{3/2}}}$

(b)

$\left.{\frac {dB(x)}{dx}}\right|_{x=R/2}=\left.-{\frac {3\mu _{0}NIR^{2}x}{2\left[R^{2}+x^{2}\right]^{5/2}}}-{\frac {3\mu _{0}NIR^{2}(x-R)}{2\left[R^{2}+(x-R)^{2}\right]^{5/2}}}\right|_{x=R/2}=-{\frac {3\mu _{0}NIR^{3}}{4\left[R^{2}+R^{2}/4\right]^{5/2}}}-{\frac {-3\mu _{0}NIR^{3}}{4\left[R^{2}+R^{2}/4\right]^{5/2}}}=0$

$\left.{\frac {d^{2}B(x)}{dx^{2}}}\right|_{x=R/2}=\left.-{\frac {3\mu _{0}NIR^{2}}{2\left[R^{2}+x^{2}\right]^{5/2}}}-{\frac {15\mu _{0}NIR^{2}x^{2}}{2\left[R^{2}+x^{2}\right]^{7/2}}}-{\frac {3\mu _{0}NIR^{2}}{2\left[R^{2}+(x-R)^{2}\right]^{5/2}}}-{\frac {15\mu _{0}NIR^{2}(x-R)^{2}}{2\left[R^{2}+(x-R)^{2}\right]^{7/2}}}\right|_{x=R/2}$ $=-{\frac {3\mu _{0}NIR^{2}}{2\left[5R^{2}/4\right]^{5/2}}}+{\frac {15\mu _{0}NIR^{4}}{8\left[5R^{2}/4\right]^{7/2}}}-{\frac {3\mu _{0}NIR^{2}}{2\left[5R^{2}/4\right]^{5/2}}}+{\frac {15\mu _{0}NIR^{4}}{8\left[5R^{2}/4\right]^{7/2}}}$ $={\frac {\mu _{0}NIR^{2}}{\left[5R^{2}/4\right]^{5/2}}}\left(-{\frac {3}{2}}+{\frac {15}{8}}{\frac {4}{5}}-{\frac {3}{2}}+{\frac {15}{8}}{\frac {4}{5}}\right)=0$

(c)

$B(0.5R)={\frac {(4\pi \times 10^{-7}{\textrm {T.m/A}})(250)(2.0{\textrm {A}})(0.10{\textrm {m}})^{2}}{\left[(0.10{\textrm {m}})^{2}+(0.05{\textrm {m}})^{2}\right]^{3/2}}}=4.5{\textrm {mT}}$